duality theorems for simplicial complexes

F A C U L T Y O F S C I E N C E U N I V E R S I T Y O F C O P E N H A G E N

Bachelor thesis in mathematicsJens Jakob Kjær

Duality theorems for simplicial complexes

Advisor: Jesper Michael Møller

June 4th, 2011

Department of Mathematical Sciences, University of Copenhagen

Abstract

This thesis deals with the homology and cohomology groups of simplicialcomplexes, and especially with the two duality theorems which willdemon-strate links between the two notions. For this we will need some results,and we will therefore prove the long exact sequence for the reduced relativehomology group; and to link the homology and cohomolgy to our geometricnotion of a simplicial complex we will see that these are connected to theEuler characteristic of the complex. Then by rewriting the relative homol-ogy group we will be able to prove the Alexander duality which is the majortheorem of this thesis. We will use the result that if realizations of twodifferent simplicial complexes are homotopic then their homology and coho-mology groups are isomorphic in order to prove another duality, by showingthat there is a different complex constructed by the nonfaces of our complexwhich is homotopic to the Alexander dual of the complex. For this we shallneed some homotopy theorems which will be duly proven.

Resume

Dette projekt beskæftiger sig med homologi- og kohomologigrupperne afsimplicielle komplekser, og specielt med to dualitets sætninger som vil viseforbindelsen mellem disse to strukturer. For at gøre dette vil vi beviseden lange eksakte sekvens for den reducerede relative homologigruppe, ogfor at forbinde homologi med kohomologi til vores geometriske intuition forsimplicielle komplekser vil vi vise, at begge begreber er forbundet til Euler-karakteristikken af et kompleks. Vi vil sa ved at omskrive pa den relativehomologi gruppe være i stand til at vise Alexander dualiteten, som er hov-edsætningen i dette projekt. Vi vil benytte resultatet, at hvis realisationeraf forskellige simplicielle komplekser er homotope da implicerer det, at dereshomologi- og kohomologigrupper er isomorfe, til at vise en anden dualitet,ved at vise at der er et kompleks konstrueret af ikke simplekser fra voreskompleks som er homotopt til det Alexander duale. Til dette skal vi benyttenogle homotopi sætninger, som vi sa vil bevise.

Contents

1 Preliminaries 21.1 Preface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.2 Acknowledgements . . . . . . . . . . . . . . . . . . . . . . . . 31.3 Simplicial Complexes . . . . . . . . . . . . . . . . . . . . . . . 3

1.3.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . 31.3.2 Duality . . . . . . . . . . . . . . . . . . . . . . . . . . 5

2 Homology and chain complexes 72.1 Chain complex . . . . . . . . . . . . . . . . . . . . . . . . . . 72.2 Homology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

2.2.1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . 82.2.2 The reduced homology . . . . . . . . . . . . . . . . . . 9

2.3 Cohomology . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142.3.1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . 142.3.2 The reduced cohomology . . . . . . . . . . . . . . . . 15

2.4 Relative homology . . . . . . . . . . . . . . . . . . . . . . . . 152.4.1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . 152.4.2 The reduced relative homology . . . . . . . . . . . . . 16

3 The Alexander duality 213.1 The Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

3.1.1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . 213.1.2 The proof . . . . . . . . . . . . . . . . . . . . . . . . . 22

4 Second duality 254.1 Topological invariance . . . . . . . . . . . . . . . . . . . . . . 254.2 Homotopic theorems . . . . . . . . . . . . . . . . . . . . . . . 27

4.2.1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . 274.2.2 Combinatorical homotopy theorems . . . . . . . . . . 29

4.3 Duality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 314.3.1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . 314.3.2 Proof . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

1

Chapter 1

Preliminaries

1.1 Preface

This bachelor thesis was was started with the intent of giving a solely com-binatorial proof of the Alexander duality. My thesis advisor did not knowof such a proof but thought it possible for me to construct one. I thereforespent the first six weeks reading up on simplicial homology theory and tryingto wrap my head around the nature of the combinatorical proofs, togetherwith the more topological proofs of the Alexander duality that we knew toexist. It had been an agreement, that if at any point I felt that I wouldnot in time be able to produce the proof I would instead write my thesis onanother subject where I would be able to use the knowledge gained in myattempt at proving the duality. The day before a meeting with my advisorI decided that if I did not see more progress I would throw in the towel,and with that thought I stumpled unto [3], an article wherein Bjorner andTancer gave a short and rather elegant proof for the duality. Bjorners proofused several of the ideas I myself halfbaked had tried. I then spent the nextcouple of weeks on writing both the proof and the definitions and resultsnecessary for the proof to work. This is what appears as chapters one tillthree. Hereafter followed the search for what would make up the rest of thethesis. First my advisor and I tried to find a connection between the vertexcoloring of a simplicial complex and its dual, but we found none, and aftera week or so we gave up. But the minor detour ended up in my readinga wide array of articles on vertex colorings. In the end I found an article,[2], with the apt name ”Note on a Combinatorial Application of AlexanderDuality”. The main result of [2] was related to the µ-operator on posets, butit also had a theorem which in this thesis is called, rather unimaginatively,the second duality. Again I needed some more results and these appear aschapters four and five of my thesis.

I have tried to include the results necerssary for my main theorems, butas always one has to stop somewhere. I have therefore chosen to consider

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1.2. ACKNOWLEDGEMENTS

as well known results and defintions of courses I have taken. These mostlyinclude algebra and topology, but there is also at least one result from linearalgebra. Likewise I have excluded the proof of the topological invariance ofboth homology and cohomology groups and instead I have only sketched.This I was advised to do, since they are rather long and technical and areso well known that it should be unthinkable that I should not see it proven.Lastly I will not for any specific space prove that it is contractible, since thistend to be a tad technical and is not very combinatorial which is the focusof the thesis. Instead I will rely on some topological intuition.

1.2 Acknowledgements

Most of the theorems, definitions and proofs presented throughout the thesisare at least inspired from other sources, and I’m most grateful to the authorsof these. For most results I will mention their source, so that a reader cangive credit where credit is due. Therefore at the beginning of each sectionI will state my major inspirational source and also note it if some result isfrom other sources.

1.3 Simplicial Complexes

This section is mostly written with help from Jesper Michael Møllers hand-written notes, but I have also used [4] and [6]. Lemma 1.8 is completly ofmy own design.

1.3.1 Definition

We wish to formally define the simplicial complexes to which most of therest of the thesis will be devoted. We want to somehow give combinatorialsets somehow equivalent to our notion of structures composed of generalizedtriangles. It is quite clear that in two dimensions such a structure is a trianglein the normal sense, composed of the convex hull of three points in the plane.So we wish to somehow identify the triangle with the set consisting of thethree vertices. Now it is also intuitively relatively clear that a line segment isa triangle-like structure in one dimension and that the triangle in the planesomehow consist of three line segments i.e. three ”triangles” of dimensionone lower. These lines egments seems to be the convex hull of two pointsin the plane, so we could identify them with the set of these two points.Now it seems clear that whenever I have one of these sets which I think ofas triangle structures all subsets seem to be the triangle structures of whichthe original structure is composed of. This line of thoughts leads us to thefollowing definition

3

1.3. SIMPLICIAL COMPLEXES

Definition 1.1. Given a finite ordered set V , we will say that K is a sim-plicial complex on the vertex set V , if K is a subset of the powerset of V ,and that if whenever σ ∈ K, and τ ⊂ σ then τ ∈ K. We will call elementsof K simplices or faces.

This raises some issues. Firstly we notice that we require V to be ordered.This doesn’t seem to stem from our geometric intuition, but it will proveitself quite useful later, when we will be able to talk about the i’the vertex ofa simplex. Secondaryly we see that if a simplicial complex K is non-emptythen ∅ ∈ K, and we will recognize the difference between ∅ as a simplicialcomplex and {∅} as a simplicial complex. We will be needing some notationfor the rest of the thesis and hence:

Definition 1.2. • The set N+ for N ∈ N0 is the set {0, 1, . . . , N} or-dered by the ”<”-relation.

• For a finite set V , then D[V ] is the power set of V viewed as a sim-plicial complex.

• For a simplicial complex K on the vertex set V , with a simplex σ, wewrite σ = {v0, v1, . . . , vn} in an unique way due to the ordering, sothat v0 < v1 < . . . < vn.

• Let B be a partially ordered by ≺, then if B is finite we will define theorder complex ∆(B) to be the simplicial complex where if

b0 ≺ b1 ≺ . . . ≺ bn (1.1)

is a chain in B then {b0, b1, . . . , bn} ∈ ∆(B). Its vertex set will be Bordered in such a way that it preserves the chains of B.

• If K is a simplicial complex then Kn, is the set of all faces σ of K,such that |σ| − 1 = n.

• If K is a simplicial complex then a subcomplex is a simplicial complexL such that L ⊂ K.

Definition 1.3. For a simplicial complex K, then we define the reducedEuler characteristic as

χ(K) =∑i

(−1)i|Ki|. (1.2)

Clearly this is a finite sum since K was finite, and it is trivially linkedto the classic notion of Vertics - Edges + Faces.

We will also need some special maps between simplicial complexes:

4


Definition 1.4. Let K,L be simplicial complexes on the vertex sets VK , VL.Then a function f : VK → VL is said to extend to a simplicial mapg : K → L if ∀σ ∈ K, with σ = {v0, . . . , vn} then {f(v0), . . . , f(vn)} ∈ L;and then we define g(σ) = {f(v0), . . . , f(vn)}. We say that two simplicialcomplexes are simplicial-isomorphic if g is bijective.

It is clear that the extended simplicial map g is bijective if and only if fis.

1.3.2 Duality

We will here introduce the notion of the Alexander dual of a simplicialcomplex. This is defined as all the complements in the vertex set to non-faces of the complex it self. This seems a rather odd definition, and indeedit is, since it doesn’t seem to be linked to any geometric intuition. But itwill still turn out most useful, since it is in the statement of the Alexanderduality.

Definition 1.5. For a simplicial complex K, on the vertex set V we definethe Alexander dual of K, K∨, to be

K∨ = {V − σ|V ⊃ σ /∈ K} (1.3)

Now we would wish that K∨ would be a simplicial complex as well, andindeed this holds:

Lemma 1.6. If K is a simplicial complex on the vertex set V , then K∨ isa simplicial complex on the vertex set V

Proof. It is trivial to see that all elements of K∨ are subset of V and henceelements of the powerset and therefore ifK∨ is a simplicial complex, it can beseen as having V as a vertex set. Suppose there is τ ⊂ σ ∈ K∨, now we knowthere is V ⊃ γ /∈ K, such that σ = V − γ, then γ ∪ (σ− τ) is not in K, sinceγ is not, and K is stable under subsets, and hence V − (γ ∪ (σ − τ)) ∈ K∨,but V − (γ ∪ (σ− τ)) = ((V − γ)−σ)∪ τ = τ , and hence K∨ is stable underintersection and therefore is a simplicial complex.

We would also very much prefer if our newfound notion was self dual,meaning that (K∨)∨ = K, and again this is a true statement, but to proveit, we will need a bit of help:

Lemma 1.7. Let K be a simplicial complex on the vertex set V , then

K∨ = D[V ]− {V − σ|σ ∈ K}. (1.4)

5


Proof. We will need to prove the two inclusions, and we therefore start bytaking an arbitrary τ ∈ D[V ]−{V −σ|σ ∈ K}, then we know that V −τ /∈ K,since if it was, then τ would not be in the set, and hence V − (V − τ) ∈ K∨and hence τ ∈ K∨. For the other inclusion we take arbitrary τ ∈ K∨, thenthere is V ⊃ σ /∈ K, such that τ = V − σ, but trivially σ ∈ D[V ], and thereis no γ ∈ K, such that V − γ = τ , and hence τ ∈ D[V ] − {V − σ|σ ∈ K},and hence we are done.

Lemma 1.8. Let K be a simplcial complex on a vertex set V , then

(K∨)∨ = K. (1.5)

Proof. We can use the lemma above to write

(K∨)∨ = D[V ]− {V − σ|σ ∈ K∨}. (1.6)

Take σ /∈ K, then V − σ ∈ K∨, and hence V − (V − σ) /∈ (K∨)∨, andtherefore (K∨)∨ ⊂ K. Now take σ /∈ (K∨)∨, then there is τ ∈ K∨ suchthat σ = V − τ , but since τ ∈ K∨ there is γ /∈ K such that V − γ = τ , butσ = (V − (V − γ)) = γ /∈ K, and hence we are done.

6

Chapter 2

Homology and chaincomplexes

2.1 Chain complex

This section is written with inspiration from Jesper Michael Møllers hand-written notes, and from [4] and [6]. Theorem 2.2 is of my own design.

Let Gn for n ∈ Z be groups and fn : Gn → Gn−1 be homomorphisms,then

(G, f) := . . .fn+1→ Gn

fn→ Gn−1fn−1→ . . . (2.1)

is a chain complex if Imfn+1 ⊂ ker fn, for all n. We say that (G, f) is exactif Imfn+1 = ker fn.

Definition 2.1. The homology of a chain complex (G, f) is the sequence ofgroups Hn(G) := ker fn/Im fn+1

It is clear that the homology group somehow measures how close a chaincomplex is to being exact. It is also clear that given two chain complexes(G, f) and (G′, f ′), and homomorphisms φn : G→ G′ such that φn−1 ◦fn =f ′n ◦ φn then there is φHn : Hn(G) → Hn(G′) given by for a z ∈ ker fn thenφHn ([z]) = [φ(z)]. This is clearly well defined since

f ′n(φn(z)) = φn−1(fn(z)) = φn−1(0) = 0 (2.2)

it is clear that if all φ are isomorphisms then H(G) ∼= H(G′) since φn canbe restricted to an isomorphism from ker fn → ker f ′n and likewise for theimages.

We will be using the following result a lot in the rest of the thesis sinceit greatly eases the definition of homomorphisms

Theorem 2.2. If G is a free group with the set of generators g = {g0, . . . , gn},H an abelian group, then every homomorphism from G to H is uniquely de-termined by the values it takes on g

7

2.2. HOMOLOGY

Proof. Let f : G→ H be a homomorphism, then take an arbitrary elementin G, it can be written on the form a0g0 + . . . + angn, for a0, . . . , an ∈ Z,then f(a0g0 + . . .+ angn) = a0f(g0) + . . .+ anf(gn).

This means that whenever we are defining homomorphisms between freegroups it is enough for us to define how they act on the set of generators.

We will wish to define the dual of a chain complex since this will beneeded twice below, in the definition of the rank of a finitely generatedabelian group and in the definition of cohomology.

Definition 2.3. Let Gn be a group and let H be a group, then Hom(Gn, H)is the set of homomorphies from Gn to H. This is a group with the usualfunction addition.

Let (G, f) be a chain complex, we can now define functions

fn : Hom(Gn−1, H)→ Hom(Gn, H) (2.3)

given as follows: For ψ ∈ Hom(Gn−1, H) and g ∈ Gn thenfn(ψ)(g) = ψ(fn(g)). It is clear from this that

(fn+1 ◦ fn)(ψ)(g) = fn+1(ψ(fn(g))) = ψ(fn(fn+1(g))) = ψ(0) = 0 (2.4)

so we see that

. . .fn→ Hom(Gn, H)

fn+1→ Hom(Gn+1, H)fn+2→ . . . (2.5)

is a chain complex.

2.2 Homology

This section is written with help from [4], where the definition of homologyis stated. The proof of theorem 2.9 is inspired by a proof strategy found insection 3.4 of [5], but is proved by me. Likewise lemma 2.6 is only stated in[4], and the proof is adapted from a similar one from [6] with some addedinputs from Sune Kristen Jakobsen and espicially Kristian Knudsen Olesen.The fact that the homology group is stable under different orderings of thevertex set, is my own proof.

2.2.1 Preliminaries

Let K be a simplicial complex. Then for all n ∈ Z we can construct theabelian group ZKn, given in the natural way

ZKn := {p0σ0 + . . .+ pmσm|p0, . . . , pm ∈ Z, σ0, . . . , σm ∈ Kn} (2.6)

8

2.2. HOMOLOGY

with the natural addition

(p0σ0 + . . .+ pmσm) + (p′0σ0 + . . .+ p′mσm)

= (p0 + p′0)σ0 + . . .+ (pm + p′m)σm (2.7)

We obviously identify Z∅ with the trivial group, which we call 0, and Z{∅}with Z.

We now wish to define functions ∂n : ZKn → ZKn−1, such that weget a chain complex from our simplicial complex. Since simplicial com-plexes are stable under subsets, it is clear that if we take a basis element{v0, . . . , vn} ∈ Kn then if we remove a single vertex from the simlpexwe get a member of Kn−1. So we define ∂n : ZKn → ZKn−1 by for{v0, v1, . . . , vn} = σ ∈ Kn

∂n(σ) =∑vi∈σ

(−1)i(σ − {vi}) (2.8)

By the argument above this is well defined, and further more if Kn = ∅,or Kn = {∅} then this clearly still works. We now wish to make sure that(ZK, ∂) is a chain complex, so we need to check that for arbitrary σ ∈ Kn,given as above, then (∂n−1 ◦ ∂n)(σ) = 0. So

(∂n−1 ◦ ∂n)(σ) = ∂n−1(∑vi∈σ

(−1)i(σ − {vi})) (2.9)

=∑vi∈σ

∑vj∈σ,j<i

(−1)i(−1)j(σ − {vi} − {vj}) (2.10)

+∑vi∈σ

∑vj∈σ,j>i

(−1)i(−1)j−1(σ − {vi} − {vj}) (2.11)

= 0 (2.12)

So from this it follows that

. . .∂n+1→ ZKn

∂n→ ZKn−1∂n−1→ . . . (2.13)

is a chain complex, which leads us to the main definition of this section:

2.2.2 The reduced homology

Definition 2.4. Let K be a simplicial complex and let ∂ be given as abovethen we call Hn(ZK) the n’th reduced homology group of K, written asHn(K) i.e. Hn(K) = ker ∂n/Im∂n−1

Take a simplicial complex K on the vertex set V , and let V ′ be V witha different ordering, now we would very much like that homology didn’tdepend on the ordering of V . Now it is clear that a new ordering of a set can

9

2.2. HOMOLOGY

be seen as a permutation of the set, and all permutations can be composed ofneighbour-switches. So we need to show that if K is a simplicial complex andwe alter the ordering on two elements so they now come in the other orderHn(K) stays the same. Take an element σ = {v0, . . . , vi, vi+1, . . . , vn} ∈ Kn,then

∂n(σ) =∑vj∈σ

(−1)jσ − {vj} (2.14)

and let σ′ = {v0, . . . , vi+1, vi, . . . , vn}, then

∂n(σ′) =∑vj∈σ′

(−1)jσ′ − {vj} (2.15)

Now it is easy to see that all generator elements of ∂n(σ) without both of theswitched vertices vi, vi+1, have the opposite sign in ∂n(σ′). And generatorelements with both switched elements keep its sign. Let φn : ZKn → ZKn,where the second Kn is ordered with the switched vertices, be given forσ ∈ Kn by

φn(σ) =

{σ if σ contains both the switched vertices−σ else

(2.16)

Now it is clear that this commutes for all n

. . .∂n+1→ ZKn

∂n→ ZKn−1∂n−1→ . . .

φn ↓ φn−1 ↓

. . .∂n+1→ ZKn

∂n→ ZKn−1∂n−1→ . . .

and hence Hn(K) is independent of the ordering of K. It is also clear thatif K,L are simplical-isomorphic then Hn(K) = Hn(L).

We will now give a result to demonstrate the usefulness of the homol-ogy definition. We want to link the reduced homology groups to the moregeometric notion of reduced Euler characteristic. But for this we will needa short result, which will come in handy in the section on cohomology aswell. This result and the following will require the notion of rank of a groupgenerated by finitely many abelian groups, so

Definition 2.5. Let A be a finitely generated abelian group, then the rankof A is defined to be

rank dimQ Hom(A,Q) (2.17)

It is clear that this is well defined since Hom(A,Q), can be seen asa vector space over the rationals, with the normal function addition andscalar multiplication. The helpful lemma is then

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2.2. HOMOLOGY

Lemma 2.6. If 0 → AfA→ B

fB→ C → 0 is exact, and A,B,C are finitelygenerated abelian groups then rankB = rankA+ rankC

Proof. So let 0 → AfA→ B

fB→ C → 0 be exact, then it is clear that fA isinjective since its kernel is trivial. We now wish to show that

0→ Hom(C,Q)fB→ Hom(B,Q)

fA→ Hom(A,Q)→ 0 (2.18)

is exact. Take arbitary ψ ∈ Hom(C,Q) such that fB(ψ) = 0, meaningthat ∀b ∈ B, we have ψ(fB(b)) = 0, but since fB is surjective it maps to allelements of C, and hence ψ = 0, so fB is injective. Now take ψ ∈ Hom(B,Q)such that fA(ψ) = 0, meaning that ψ ◦ fA = 0 hence we can induce ahomomorphism ψ′ : B/fA(A)→ Q, we can likewise induce an isomorphismfrom fB f ′B : B/fA(A) → C, since the original chain was exact, now letφ : C → Q, be given by φ = ψ′◦(f ′B)−1, then we clearly see that fB(φ) = ψ.Now take an arbitary ψ ∈ Hom(A,Q), we define a set X, where elementsof X, are on the form (H,φ), where fA(A) ⊂ H ⊂ B, with H a group andφ a homomorphism from H to Q.It is now clear that X is non-empty since(fA(A), ψ ◦f−1

A ) ∈ X, since fA is injective. We partially order X, by writingthat

(H,φ) ≤ (H ′, φ′)⇔ H ⊂ H ′ and φ = φ′|H (2.19)

Now for an arbitary chain in X

(H1, φ1) ≤ (H2, φ2) ≤ . . . (2.20)

Then clearly H =⋃∞i=1Hi is a subgroup of B and φ : H → Q is a homo-

morphism given as follows: For x ∈ H, then φ(x) = φn(x) for x ∈ Hn. Andhence (H,φ) ∈ X, and for all i ∈ N then (Hi, φi) ≤ (H, φ). We can now useZorns lemma to state that there is a maximal element (H, φ) ∈ X, and wewish to show that H = B. Assume for contradiction that there is b ∈ B−H,then clearly the set of k ∈ Z such that k ·b ∈ H is an ideal in Z, and since Z isa prime ideal domain there is k0 ∈ Z such that (k0) is this ideal. Now clearly

we can pick y ∈ Q such that φ(k0·b) = y·k0. Now take the group K =⟨H, b

⟩being the subgroup of B generated by H and b. Now define β : K → Qgiven as: For h ∈ H and n ∈ Z β(h+n·b) = φ(h)+y ·k0. Now this is well de-fined since if we take n ∈ Z such that n ·b ∈ H, but then there is m ∈ Z suchthat n = mk0, then β(nb) = φ(nb) = φ(mk0b) = mφ(k0b) = myk0b = nb.So β is a homomorphism, and β|H = φ and hence (H, φ) ≤ (K,β), but this

is a contradiction and hence H = K, and therefore φ ∈ Hom(B,Q) suchthat fA(φ) = ψ. And now we know that

0→ Hom(C,Q)fB→ Hom(B,Q)

fA→ Hom(A,Q)→ 0 (2.21)

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2.2. HOMOLOGY

is exact. It is trivial to see that fA and fA can be seen as linear functionsand hence we can now use the dimension theorem of linear algebra and statethat

rankB = dimQ Hom(B,Q = rgfA + dimQ ker fA (2.22)

= dimQ ImfA + dimQ ImfB = dimQ Hom(A,Q) + dimQ Hom(C,Q)

= rankA+ rankC (2.23)

And hence we are done.

Theorem 2.7. For a simplicial complex K, then

χ(K) =∑i

(−1)irankHi(K). (2.24)

Proof. We wish to show that with i as the inclusion map

0→ ker ∂ni→ ZKn

∂n→ Im∂n → 0 (2.25)

is exact. This is clear since i is injective so its kernel is just 0, and the kernelof ∂n is by definition ker ∂n, and the kernel of something which maps to thetrivial group is the group itself and hence it is exact, so by lemma 2.6 wesee that

rankZKn = rank ker ∂n + rank Im∂n (2.26)

Likewise we wish to convince ourselves that

0→ Im∂n+1i→ ker ∂n

q→ Hn → 0 (2.27)

, with i as the inclusion map, which makes sense since ∂ is a chain map andq as the quotient map, is exact. Again since i is injective its kernel is 0,and the kernel of the quotient map is the thing you mod away, in this caseIm∂n+1, and since q is surjective its image is all of Hn(K), which is exactlywhat gets mapped to 0, so this is exact as well. Now we want to know therank of ZKn, but since any homomorphism between ZKn and Q is going tobe uniquely determined by its values on elements of Kn, then we triviallysee that the rank is equal to the cardinality. Combining lemma 2.6 with theresults above

|Kn| = rankIm∂n+1 + rankIm∂n + rankHn(K) (2.28)

And this gives that

χ(K) =∑i

(−1)irankIm∂n+1 + rankIm∂n + rankHn(K) (2.29)

= −rankIm∂−1 + (−1)N rankIm∂N+1 +

N∑i=−1

Hi(K) (2.30)

where N ∈ N0 such that Kj = ∅ for all j ≥ N . But obviouslyrankIm∂−1 = rankIm∂N+1 = 0 and hence we are done.

12

2.2. HOMOLOGY

Another result about the reduced homology which we will need later inthe thesis is that the full simplicial complex (the complex which includes thevertex set as a simplex) has the trivial reduced homology group for all n.

Definition 2.8. A simplicial complex K is called acyclic if Hn(K) = 0, forall n ∈ Z

With this terminology we want to prove that

Lemma 2.9. D[N+] is acyclic for all N ∈ N0

Proof. We will prove this by induction, and therefore let N = 0, then triv-ially Hn(D[0+]) = 0 since the chain complex is . . .→ 0→ Z{0} → Z→ 0→. . .. Now assume thatD[M+] is acyclic for allM ≤ N . Take z ∈ ZD[N+1+]nsuch that ∂n(z) = 0, we know that z = p0σ0 + . . . + pmσm, but for each σjeither σj ∈ D[N+]n or there is some τj ∈ D[N+]n−1 such that

σj = τj ∪ {N + 1} (2.31)

So we can now rewrite z as

z = a0σ0 + . . .+ atσt + at+1τ1 ∪ {N + 1}+ . . .+ at+sτs ∪ {N + 1} (2.32)

for a0, . . . , at+s ∈ Z, σ0, . . . , σt ∈ D[N+]n and τ1, . . . , τs ∈ D[N+]n−1. We seethat ∂n(τj∪{N+1}) = ∂n−1(τj)∪{N+1}+(−1)nτj , where ∂n−1(τj)∪{N+1}means that we add N + 1 to each simplex we get from ∂. We see that

0 = ∂n(z)

= a0∂n(σ0) + . . .+ at∂n(σt) + at+1∂n(τ1 ∪ {N + 1}) + . . .+ at+s∂n(τs ∪ {N + 1})= ∂n(a0σ0 + . . .+ atσt) + ∂n−1(at+1τ1 + . . .+ at+sτs) ∪ {N + 1}

+ (−1)nat+1τ1 + . . .+ at+sτs

From this it is clear that ∂n−1(at+1τ1 + . . .+at+sτs)∪{N+1} = 0 since theseare the only elements which contain N + 1 as an element in the simplexes.But if that is true then it must follow that ∂n−1(at+1τ1 + . . .+ at+sτs) = 0,and since D[N+] is acyclic its chain complex must be exact, and hence theremust be an x ∈ D[N+]n such that ∂n(x) = at+1τ1 + . . . + at+sτs. But withthis we find that

0 = ∂nZ

= a0∂n(σ0) + . . .+ at∂n(σt) + at+1∂n(τ1 ∪ {N + 1}) + . . .+ at+s∂n(τs ∪ {N + 1})= a0∂n(σ0) + . . .+ at∂n(σt) + (−1)nat+1τ1 + . . .+ at+sτs

= a0∂n(σ0) + . . .+ at∂n(σt) + (−1)n∂n(x) (2.33)

But a0σ0 + . . . + atσt + (−1)nx ∈ D[N+]n which is acyclic and hence thereis y ∈ D[N+]n+1 such that ∂n+1(y) = a0σ0 + . . .+atσt+ (−1)nx, and we see

13

2.3. COHOMOLOGY

that y + x ∪ {N + 1} ∈ D[N + 1+]n+1, and that

∂n+1(y + x ∪ {N + 1}) = ∂n+1(y) + ∂n+1(x ∪ {N + 1}) (2.34)

= ∂n+1(y) + ∂n(x) ∪ {N + 1}+ (−1)n+1x (2.35)

= a0σ0 + . . .+ atσt + (−1)nx+ ∂n(x) ∪ {N + 1}+ (−1)n+1x

= a0σ0 + . . .+ atσt + (at+1τ1 + . . .+ at+sτs) ∪ {N + 1}= z (2.36)

And hence we see that for an arbitrary [z] ∈ Hn(D[N + 1+]), then

[z] = [∂n+1(y + x ∪ {N + 1})] = [0] (2.37)

, and hence D[N + 1+] is acyclic, and this concludes the proof.

This lemma might seem at the moment somewhat arbitrary, but it willcome in most handy when we prove the Alexander duality. However thiswill require a great deal more theory, so we will carry on:

2.3 Cohomology

This section too is written with [4]. Theorem 2.11 is stated without proofin [8].

2.3.1 Preliminaries

Again we will now turn our attention to the group of homomorphisms froma group to a given set, this time the set of integers. We wish to againconstruct a chain complex somehow related to our simplicial complex. Thistime we want it to be the dual of the homology so to speak. As we have seenearlier if one has homomorphisms between groups it is quite easy to constructhomeomorphisms between the set of homomophisms into a given group. Solet ∂n be given as in the previous section, and let K be a simplicial complex,then define δn : Hom(ZKn−1,Z)→ Hom(ZKn,Z), given as follows: For anarbitrary homomorphy ψ : ZKn−1 → Z, then δn(ψ) = ψ◦∂n. This is clearlywell defined since both ∂ and ψ are homomorphisms and hence so is theircomposition. Then clearly we find that

(δn+1 ◦ δn)(ψ) = δn+1(ψ ◦ ∂n) = (ψ ◦ ∂n) ◦ ∂n+1 = ψ ◦ 0 = 0 (2.38)

with 0 being the trivial homomorphy. Then except for the numbering whichseems to go the other way than usually, we find that

. . .δn→ Hom(ZKn,Z)

δn−1→ Hom(ZKn−1,Z)δn−1→ . . . (2.39)

Is a chain complex.We can now as previously move on to a definition.

14

2.4. RELATIVE HOMOLOGY

2.3.2 The reduced cohomology

Definition 2.10. Let K be a simplicial complex, and δj given as above, thenthe n’th reduced cohomology group of the complex K is

Hn(K) := ker δn+1/Imδn. (2.40)

This definition seems very close to our reduced homology groups, andas we will see throughout the thesis this is quite true; there are severalduality theorems (one of which is the Alexander duality) linking homologyand cohomology of different objects together. Like homology cohomologylike homology is independent of the ordering of the vertex set. We will notshow this since the proof is rather similar to the argument for homology. Wewill start by showing that as homology the cohomology has the same linkto the geometric notion of a simplicial complex, since:

Theorem 2.11. For a simplicial complex K, then

χ(K) =∑i

(−1)irankH i(K). (2.41)

The proof for this is completely analogous to that for the homology caseand therefore I will omit the proof. The only thing that changes slightlyis that we would need to realize that rankHom(ZKn,Z) = |Kn|, but this isclear since a homology from a free group to Z is merely a function from theset of generators to Z.

2.4 Relative homology

This section is likewise inspired by [4]. Corollary 2.15 is from [3]. The result(2.64) is stated in [3], but proved by me.

2.4.1 Preliminaries

Our reason for studing simplicial complexes is that, given a topological spaceit is often quite easy to triangulate it and then find a simplicial complex,the homology of which we can then study. But sometimes it is not quiteclear how this triangulation should go. It is e.g. clear that a ball in threedimensions seems to be equivalent to D[3+], but what about the ball wherewe remove the sphere? It is not quite clear what simplicial complex it shouldbe. We can easily see that the sphere should be D[3+]−{0, 1, 2, 3}, but since{{0, 1, 2, 3}} isn’t a simplicial complex then our methods so far are at a loss.We could let the homology be the quotient between the homology groups,but this isn’t always well defined so instead we construct the following:

For K,A simplicial complexes with A ⊂ K then we have

q : ZKn → ZKn/ZAn (2.42)

15


being the quotient map. Then define

∂∗n : ZKn/ZAn → ZKn−1/ZAn−1 (2.43)

given by for x ∈ ZKn/ZAn then take y ∈ ZKn such that q(y) = x, andthen ∂∗n(x) = q(∂n(y)). We need to check that this is well defined, so lety, y′ ∈ ZKn such that q(y) = q(y′) = x, then there is a ∈ ZAn such thaty = y′ + a. Since A is a simplicial complex, and hence stable under subsetsit is clear that ∂n(a) ∈ ZAn−1, and hence

q(∂n(y)) = q(∂n(y′ + a)) = q(∂n(y′) + ∂n(a)) (2.44)

= q(∂n(y′)) + q(∂n(0)) = q(∂n(y′)) (2.45)

If we take arbitrary x, x′ ∈ ZKn/ZAn, with y, y′ ∈ ZKn such that q(y) = xand q(y′) = x′ then

∂∗n(x) + ∂∗n(x′) = q(∂n(y)) + q(∂n(y′)) = q(∂n(y) + ∂n(y′)) (2.46)

= q(∂n(y + y′)) = ∂∗n(x+ x′) (2.47)

Thus ∂∗n is a homomorphism. Since we wish to construct a chain complexwe need to check that the image is in the kernel for our new function so letx ∈ ZKn/ZAn be arbitrary, and y ∈ ZKn be so that q(y) = x then

(∂∗n−1 ◦ ∂∗n)(x) = ∂∗n−1(q(∂n(y))) = q((∂n−1 ◦ ∂n)(y)) = q(0) = 0 (2.48)

We are now ready to define our new notion of relative homology:

2.4.2 The reduced relative homology

Definition 2.12. Let K,A be simplicial complexes with A ⊂ K and let ∂∗nbe given as above, then the n’th reduced relative homology group between Kand A is Hn(K,A) = ker ∂∗n/Im∂∗n+1.

Again we will abstain from showing that this group is independent of theordering of the vertex set. Again the argument would be rather similar tothe that for homology case. This relative homology tends to come in handyespecially for calculating various homology groups, since the next theoremwe will show gives an exact sequence linking the reduced homology groupsof K and A to the their relative homology. For this we will need a functionϕ : Hn(K,A)→ Hn−1(A), so we will construct that.

Remark 2.13. from this point on we will stop indexing our functions ∂n, ∂∗n

and δn since it is very seldom that the indices do anything but hinder boththe imagination and notation of proofs and constructions. We will howeverinclude the index if somewhere it should be important.

16


Consider the following diagram

. . .∂→ ZAn+1

∂→ ZAn∂→ ZAn−1

∂→ . . .i ↓ i ↓ i ↓

. . .∂→ ZKn+1

∂→ ZKn∂→ ZKn−1

∂→ . . .q ↓ q ↓ q ↓

. . .∂∗→ ZKn+1/ZAn+1

∂∗→ ZKn/ZAn∂∗→ ZKn−1/ZAn−1

∂∗→ . . .

where the rows are chain complexes, i is the inclusion map, and q the quo-tient map. It is rather trivial that i ◦ ∂ = ∂ ◦ i, and likewise from thedefinition that ∂∗ ◦ q = q ◦ ∂, and hence the diagram commutes which is arather useful fact, for then we can define iH and qH as we did in section 2.1.

Theorem 2.14. Long exact sequence: If K,A are simplicial complexeswith A ⊂ K then the following is an exact sequence for the functions givenabove

. . .ϕ→ Hn(A)

iH→ Hn(K)qH→ Hn(K,A)

ϕ→ Hn−1(A)iH→ . . . (2.49)

The following proof is a diagram chase, and hence it is most useful tokeep the diagram above in mind as one reads it.

Proof. We will do this proof one step at a time, checking one inclusion afterthe other

• ImiH ⊂ ker qH

For arbitrary [a] ∈ Hn(A), then qH ◦ iH([a]) = qH([i(a)]) = [q(i(a))] = [0],and hence we have the inclusion.

• ImqH ⊂ kerϕ

Take arbitrary [x] ∈ Hn(K), then we know that ∂(x) = 0, so if q(x) = z,then ϕ(qH([x])) = [0].

• Imϕ ⊂ ker iH

For arbitrary [z] ∈ Hn(K,A), then there exists x ∈ ZKn such that q(x) = z,but then iH(ϕ[z]) = iH([a]) = [i(a)] = [∂(x)] = 0.

• ker qH ⊂ ImiH

Fix x ∈ ker qH , then we know there is z ∈ ZKn+1/ZAn+1, such thatq(x) = ∂∗z, since q is surjectiv we get a x′ ∈ ZKn+1, such that q(x′) = z.We now see that q(x − ∂(x′)) = q(x) − ∂(q(x′)) = ∂∗(z) − ∂∗(z) = 0, sox− ∂(x′) ∈ ker q = Imi, hence ∃a ∈ ZAn such that i(a) = x− ∂(x′). We seethat ∂(i(a)) = ∂(x−∂(x′)) = ∂(x) = 0, and since i is injectiv then a ∈ ker ∂,and hence [a] ∈ Hn(A), and we find that iH([a]) = [i(a)] = [x− ∂(x′)] = [x].

17


• kerϕ ⊂ ImqH

Take [z] ∈ kerϕ then ∃a ∈ ZAn such that ϕ([z]) = [0] = [∂a], now takex ∈ ZKn such that q(x) = z and see that ∂(x− i(a)) = ∂(x)− i(∂(a)) = 0,due to the definition of the ϕ function, we find [x − i(a)] ∈ Hn(K) andqH([x− i(a)]) = [q(x)− q(i(a))] = [z], hence ker q = Im i.

• ker iH ⊂ Imϕ

Fix a ∈ ker iH , then i(a) = ∂(x), for some x ∈ ZKn, and

∂∗(q(x)) = q(∂(x)) = q(i(a)) = 0 (2.50)

so [q(x)] ∈ Hn(K,A), and from this it clearly follows that ϕ([q(x)]) = [a]. Wehave now shown all the necessary inclusions. This concludes the proof.

As promised above, the concept of the reduced relative homology is quitehandy for calculating homology groups, and here we will give a short corol-lary to show the usefulness, the result will be used in the proof of the Alexan-der duality.

Corollary 2.15. Let K be a simplicial compleks on the vertex set of V thenit follows that Hn−1(K) ∼= Hn(D[V ],K) for all n

Proof. Trivially K ⊂ D[V ], so hence the relative homology group is defined.It is also clear that we can relabel V with N+ for a suitable N ∈ N0 sinceV is finite and hence Hn(D[V ]) = 0 for all n due to lemma 2.9. If we usethis result in the long exact sequence above then we find that for all n,0 → Hn(D[V ],K)

ϕ→ Hn−1(D[V ],K) → 0 is exact. This implies that ϕ isinjectiv since its kernel must be the image of 0, hence 0. And it is surjectivesince its image must be the kernel of a homomorphy mapping into the trivialgroup, and hence we are done.

In this section we will need another small result to prove the Alexanderduality. Since later it will make our lives easier we want to invent a newnotation and prove it equivalent to the one we have already introduced. Solet A,K be simplicial complexes such that A ⊂ K then we will use thegroups Z(Kn − An) where ”−” is the set theoretical difference. We wish tomake a chain complex so we will need a function. Therefore we will defined : Z(Kn −An)→ Z(Kn −An) given as follows: For σ ∈ Kn −An then

d(σ) =∑

vi∈σ,σ−{vi}/∈A

(−1)iσ − {vi} (2.51)

18


Then we have that

d ◦ d(σ) = d

∑vi∈σ

σ−{vi}/∈A

(−1)iσ − {vi}

(2.52)

=∑vi∈σ

σ−{vi}/∈A

(−1)i∑

vj∈σ−{vi}σ−{vi,vj}/∈A

(−1)jσ − {vi, vj} (2.53)

=∑vi∈σ

σ−{vi}/∈A

∑vj∈σ,vj<viσ−{vi,vj}/∈A

(−1)i(−1)jσ − {vj , vi} (2.54)

+∑vi∈σ

σ−{vi}/∈A

∑vj∈σ,vj>viσ−{vi,vj}/∈A

(−1)i(−1)jσ − {vj , vi} (2.55)

= 0 (2.56)

So we now have a chain complex, which we wish to show to be isomorphicto the relativ homology group, so we need an isomorphism

f : Z(Kn −An)→ ZKn/ZAn (2.57)

such that ∂∗ ◦ f = f ◦ d. Now let σ ∈ Kn − An then trivially σ ∈ Kn andhence q(σ) ∈ ZKn/ZAn with q being the quotient map, so let f(σ) = q(σ).

(∂∗ ◦ f)(σ) = ∂∗(q(σ)) = q(∂(σ)) = q(∑vi∈σ

(−1)i(σ − {vi})) (2.58)

=∑vi∈σ

(−1)iq((σ − {vi})) (2.59)

=∑vi∈σ

σ−{vi}/∈A

(−1)iq((σ − {vi})) (2.60)

(f ◦ d)(σ) = f

∑vi∈σ

σ−{vi}/∈A

(−1)iσ − {vi}

(2.61)

=∑vi∈σ

σ−{vi}/∈A

(−1)iq(σ − {vi}) (2.62)

We need to show that f is an isomorphism, but clearly it is injective sinceother than elements of ZAn the only thing that maps to 0 is 0 itself, soits kernel is trivial. If we take arbitrary [x] ∈ ZKn/ZAn, then there isx′ ∈ Z(Kn −An) and a ∈ ZAn such that x′ + a = x, but then

[x] = [x′ + a] = [x′] (2.63)

19


and then f(x′) = [x], and hence f is surjective. We now see that

Hn(K,A) = ker dn/Im dn+1. (2.64)

20

Chapter 3

The Alexander duality

This entire chapter is inspired by [3], the only exception being corollary 3.3,which is stated with an alternative proof in [8], but this proof is by me.

3.1 The Theorem

3.1.1 Preliminaries

We are now ready to both state and prove the Alexander duality. As we sawin the section on cohomology there is a close link between homology andcohomology, and it is one of these connections between the two notions wewish to prove. The theorem states that

Theorem 3.1. Alexander duality: For a simplicial complex K, on avertex set V , with |V | − 1 = N the following statement holds

Hi(K) ∼= Hj(K∨) (3.1)

whenever i+ j = N − 2

There are several versions of the theorem, all being equivalent. The oneabove is the formulation I have chosen to prove.

We will start by sketching out the proof, so that when the formal proofis given below we will not loose our way. We remember that for a simplicialcomplex K the elements of K∨ are complements of non-simplexes of K.So if somehow we could construct a complement map from the set of non-simplexes it would seem doable. But we seem to already have somethingthat more or less is this, namely the relative group ZD[V ]/ZKn, whichhas a homology isomorphic to K’s reduced homology due to the long exactsequence. So we wish to map these elements to homologies from ZK∨n toZ. It is clear that every element in Hom(ZK∨n ,Z) can be described as thesum of functions 1τ : ZK∨n → Z, for τ ∈ K∨n given as follows: If x =a0σ0 + . . . + am−1σm + amτ then 1τ (x) = am. So we could hope that if we

21

3.1. THE THEOREM

map σ ∈ ZD[V ]/ZKn to 1V−σ then everything would work out nicely. Thisdoes not hold, but if we remember to do something about the sign of thefunction it will commute with δ and ∂∗ and then we will be done.

3.1.2 The proof

Proof of the Alexander duality. Firstly we will need that the group

A := {∑σ∈Kn

aσ1σ|aσ ∈ Z} (3.2)

with 1σ defined as above is equal to Hom(ZKn,Z). It is rather trivial thatA ⊂ Hom(ZKn,Z). And if we have ψ ∈ Hom(ZKn,Z), then ψ is uniquelydetermined by it actions on elements in Kn and hence we can compose it ofelements of A and therefore {

∑σ∈Kn aσ1σ|aσ ∈ Z} = Hom(ZKn,Z), since

their composition rules are trivially identical.Now let K be a simplicial complex on the vertex set V , with |V |−1 = N ,

then clearly we can rename the vertices such that K ⊂ D[N+] then we wishto define a homomorphy so that for σ ∈ (D[N+]i+1 −Ki+1) then we define

f : Z(D[N+]i+1 −Ki+1)→

∑σ∈K∨N−i−2

aσ1σ

∣∣∣∣∣∣ aσ ∈ Z

(3.3)

f : σ 7→ Πvi∈σ(−1)vi1V−σ (3.4)

Then this is well defined due to the definition of K∨. It is rather triviallya homomorphy, and it is clear that the only thing f maps to the constantfunction 0 is 0, so it is injective. Given a homomorphy we can decompose itinto a sum of aσ1σ, for suitable σ’s and aσ ∈ Z, and hence f is surjective.

22

3.1. THE THEOREM

So for σ ∈ (D[N+]i+1 −Ki+1) and τ ∈ K∨N−i−2 we find that

(φ ◦ d)(σ)(τ) = φ

∑vi∈σ,σ−{vi}/∈K

(−1)iσ − {vi}

(τ) (3.5)

=∑

vi∈σ,σ−{vi}/∈K

(−1)iΠj∈σ−{vi}(−1)j1N+−(σ−{vi})(τ) (3.6)

=∑

vi∈σ,σ−{vi}/∈K,τ=N+−(σ−{vi})

(−1)iΠj∈σ−{vi}(−1)j (3.7)

=∑

vi∈σ,τ−{vi}=N+−σ

(−1)iΠj∈σ−{vi}(−1)j (3.8)

(δ ◦ φ)(σ)(τ) = δ(Πi∈σ(−1)i1N+−σ)(τ) (3.9)

= Πi∈σ(−1)i1N+−σ(∂(τ)) (3.10)

= Πi∈σ(−1)i1N+−σ

(∑vi∈τ

(−1)iτ − {vi}

)(3.11)

= Πi∈σ(−1)i∑

vi∈τ,τ−{vi}=N+−σ

(−1)i (3.12)

It is clear that if there are no vi ∈ τ such that τ −{vi} = N+− σ then bothsums are empty and hence (φ◦d)(σ)(τ) = 0 = (δ ◦φ)(σ)(τ). So assume thatthere is vi ∈ τ such that τ − {vi} = N+ − σ, then clearly vi = vk ∈ σ forsome k, and vi is unique, then we find that

(φ ◦ d)(σ)(τ) = (−1)k ·Πj∈σ−{vk}(−1)j = Πj∈σ,j<vk(−1) ·Πj∈σ−{vk}(−1)j

(δ ◦ φ)(σ)(τ) = Πj∈σ(−1)j · (−1)i = Πj∈σ(−1)j ·Πl∈N+−σ,l<vk(−1) (3.13)

It is clear by simple arithmetic that these to products are identical, andtherefore

ker di+1/Im di+2∼= ker δN−i−1/ImδN−i−2 (3.14)

and now we are done since

Hi(K)Cor. 2.9∼= Hi+1(D[N+],K)

(2.64)∼= ker di+1/Imdi+2 (3.15)

(3.14)∼= ker δN−i−1/ImδN−i−2 = HN−i−2(K∨) (3.16)

Corollary 3.2. Let K be a simplicial complex on the vertex set V , with|V | − 1 = N . Then

Hi(K∨) ∼= Hj(K) (3.17)

whenever i+ j = N + 2

23

3.1. THE THEOREM

Proof. This trivially follows from the fact that (K∨)∨ = K and from theAlexander duality.

Corollary 3.3. Let K be a simplicial complex, with N+ as the vertex set,then χ(K) = (−1)N−2χ(K∨)

Proof. We have that χ(K) =∑

i(−1)irankHi(K) and using the Alexanderduality we have that

χ(K) =∑i

(−1)irankHi(K) =∑i

(−1)irankHN−2−i(K∨) (3.18)

=∑i

(−1)N−2−irankH i(K∨) = (−1)N−2∑i

(−1)−irankH i (3.19)

= (−1)N−2∑i

(−1)irankH i = (−1)N−2χ(K∨) (3.20)

and hence we are done.

24

Chapter 4

Second duality

4.1 Topological invariance

This section is a sketch of proofs and theorems found in [6].In this chapter, in pursuit of a different duality between homology and

cohomology, we willl try to realize our simplicial complexes as topologicalspaces. For this we will need some basic definitions.

Definition 4.1. Let {a0, a1, . . . , an} be a set of points in RN for some N ,then they are geometrically independent, if a1 − a0, . . . , an − a0 are linearlyindependent in the usual linear algebra way.

Given a set of geometrically independent points, {a0, a1, . . . , an}, theirconvex hull is the set of points

∑ni=0 tiai, where ti ∈ R and are non-negative,

and∑n

i=0 ti = 1. Given a convex hull of {a0, a1, . . . , an}, it faces are theconvex hulls of subsets of {a0, a1, . . . , an}

It is clear that for a point in a convex hull the ti’s are uniquely deter-mined. We now need a topological equivalent to our simplicial complexes,so we define.

Definition 4.2. A simplicial polytope, P is a subset of the powerset RN forsome N consisting of convex hulls of finite set of points such that

• For any convex hull σ in P then all faces of σ are also in P.

• All intersections between two elements of P are faces of both elementsof P.

We will write ‖P‖ :=⋃s∈P s, and call this object the realization of P. We

give to this space the following topology: A subset A of ‖P‖ is closed if A∩sis closed in the standard topology on RN for all s ∈ P.

Now it appears that somehow P is closely related to our simplicial com-plexes. The simplicial polytope and the simplicial complex actually seems

25

4.1. TOPOLOGICAL INVARIANCE

to be equivalent since given a simplicial complex K, then take a functionfrom f : K0 → RN such that the images of the vertices are geometricallyindependent, then let

K = {convex hull of{f(vi0), . . . , f(vin)}|{vi0 , . . . , vin} ∈ K} (4.1)

This is clearly a simplicial polytope. Given a simplicial polytope P we candefine

P = {{ai0,...,ain}|The convex hull of {ai0,...,ain} ∈ P} (4.2)

and this is clearly a simplicial complex. Throughout the following chapterwe will also view a simplicial complex as a simplicial polytope, and, e.g.,write ‖K‖ for the geometricale realization of the simplicial polytope whichcorresponds to the simplicial complex K 6= {∅}.

For simplicial complexes K,L let f : K → L, be a simplicial map, thenclearly f∗ : ‖K‖ → ‖L‖ given as follows: For an element in x ∈ ‖K‖ thenthere is a simplex in K, {a0, . . . , an} such that x in the convex hull of this,then x =

∑ni=0 tiai then f∗(x) =

∑ni=0 tif(ai). This is clearly welldefined

and continuous. It is also clear that if f was simplicial-isomorphy then‖K‖, ‖L‖ are homeomorphic.

We will now sketch the way to prove the following result:

Theorem 4.3. Let K,L be simplicial complexes such that ‖K‖ is homotopicequivalent to ‖L‖, then Hn(K) ∼= Hn(L)

Now we would wish for that if f : ‖K‖ → ‖L‖ was continuous thenwe would be able to induce a simplicial-map between K,L but this is notthe case. Then we would like, that f induces some homomorphy betweenHn(K) and Hn(L) this is not directly true either, but by going slightly outof our way we will be able to make a isomorphy to some ”subdivision” ofK’s homology group from K’s homology and then a homomorphy onwardsto the homology of L. We will need the following

Definition 4.4. Let K be a simplicial complex and let v be a vertex of K,then the st (v,K), or just st v if the complex is clear, is the subset of ‖K‖given by the union of the interiors of the simplices of K that have v as avertex.

That this is such an important definition seems surprising but it shallprove most useful.

Definition 4.5. Let K,L be simplicial complexes and let h : ‖K‖ → ‖L‖be continuous, then we say that h satisfies the star condition with respect toK and L if for each vertex {v} ∈ K there is a vertex {w} ∈ L, such thath(st v) ⊂ st w.

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4.2. HOMOTOPIC THEOREMS

It can be proven that if h : ‖K‖ → ‖L‖ satisfies the star conditionthen it induces a simplicial map K → L, but it is also clear that not allmaps satisfy the star condition. Now let the subdivision of a simplicialcomplex K be sdK = ∆(K − ∅), where the K on the right hand sight isviewed as a poset ordered by the subset-ordering. Then one could provethat for a continuous map f : ‖K‖ → ‖L‖ there exists N ∈ N0 such thatthere is a continuous map f ′ : ‖sdNK‖ → ‖L‖ such that f ′ has a simplicialapproximation from sdNK → L. Now it is provable that there is a isomorphyHn(K) → Hn(sdNK), and so for any continuous map we can think of itsinduced simplicial map, since we can merely use the simplicial map inducedby the function from some subdevision, and one can therefore show that thehomology group of a simplicial complex is stable under homeomorphisms ofits realization.

Now to prove theorem 4.3, one first needs to realize that there is a sim-plicial complex, which can be realized as the space ‖K‖× I, with I = [0, 1].Then one can show that if two continuous maps between realizations are ho-motopic then their induced homomorphisms between the homology groupsare equal. Because then if K,L realize as homotopic equivalent spaces thenthe homotopic equivalences induces inverse homomorphisms whose compo-sition is the identity i.e. they are isomorphisms.

The same process will clearly show us that likewise cohomology is stableunder homotopy. And we can now define the reduced homology and coho-mology of a topological space X, which is homotopic to some realization ofa simplicial complex K

Definition 4.6. If a topological space is homotopic to some realization ofsome simplicial complex K, then we define the n’th reduced homology groupof X: Hn(X) = Hn(K) and like wise the n’th reduced cohomology group ofX: Hn(X) = Hn(K)

4.2 Homotopic theorems

This section is inspired by [1], with the exception of lemma 4.14, which isonly stated there, but is proved in [9], and theorem 4.13 which is adaptedfrom a proof in [7].

4.2.1 Preliminaries

In this section we will prove some theorems bearing on the question: Whenare realizations of different complexes homotopic. For this we will have useof some definitions.

Definition 4.7. Let K be a simplicial complex and T a topological space.Let C : K → P(T ) such that if σ ⊂ τ for σ, τ ∈ K then C(σ) ∈ C(τ).

27


A mapping f : ‖K‖ → T is said to be carried by C if for all σ ∈ K,f(‖σ‖) ⊂ C(σ).

We will often use that maps of interest between simplicial complexeshas a continous realization which is carried by something mapping to acontractible space.

Definition 4.8. Let K be a simplicial complex and P a poset ordered by≺ then f : K → P is po-simplicial if σ ∈ K, with σ = {v0, . . . , vn}, thenf(vi0) ≺ . . . ≺ f(vin), where vij is an unique element in {v0, . . . , vn}.

In other words a po-simplicial function maps simplices to chains.

Definition 4.9. Given a family of sets (Ai)i∈I indexed by a finite orderedset I, the nerve of the family is the simplicial complex N (Ai) defined on thevertex set I, such that σ ∈ N (Ai) if σ ⊂ I and

⋂i∈σ Ai 6= ∅, where we define⋂

i∈∅Ai :=⋃i∈I Ai

It is clear that this is a simplicial complex since, if σ ∈ N (Ai) and τ ⊂ σthen

⋂i∈σ Ai ⊂

⋂i∈τ Ai, so τ ∈ N (Ai).

Definition 4.10. Let P be a poset ordered by ≺, a subset C is called acrosscut if

1. no two elements in C are comparable

2. for every finite chain a0 ≺ . . . ≺ an, with a0, . . . , an ∈ P , there is somea ∈ C such that a is comparable to each ai.

3. for all A ⊂ C for which there is an element p ∈ P such that a ≺ p forall a ∈ A, then there should be a unique least upper bound for A in P .Like wise if there is p ∈ P such that p ≺ a for all a ∈ A, then thereshould be a unique greatest lower bound for A in P .

Definition 4.11. If C is a finite crosscut then Γ(P,C) determines a sim-plicial complex consisting of all subsets of C such that there is an upper ora lower bound in P

We will also be needing a more topological result and for that a definition

Definition 4.12. A cell is a topological space homeomorphic with the unitball Bn for some n ∈ N0

Theorem 4.13. If X is a cell, and T is a contractible space, then if f isa continuous function from the boundary of X to T , then f extends to acontinuous function f : X → T .

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Proof. For X being homeomorph with B0 this is a trivial statement sincethen X consists of a single point, and hence is equal to its own bound-ary. Next let n ≥ 1. Let h : X → Bn be the homeomorphism, andlet F be the homotopy between idT and the constant map. Now clearlyh maps the boundary of X to Sn−1. Now define H : Sn−1 × I → Tby H(x, t) = F ((f ◦ h−1)(x), t), then clearly this is a homotopy betweenf ◦ h−1 and the constant map. Now let q : Sn−1 × I → Bn, be given byq(x, t) = (1− t)x, clearly this is continuous, closed and surjective, so it is aquotient map collapsing Sn−1×{1} to 0. Since H is constant on Sn−1×{1},then clearly it induces a continuous function k : Bn → T via the quotientmap. Now let f = h−1 ◦ k, then it is clear that for any element x in theboundary of X, it holds that f(x) = f(x).

4.2.2 Combinatorical homotopy theorems

We will be needing a lemma for use in the following theorem, namely:

Lemma 4.14. Let K be a simplicial complex, let T a topological space andlet C : K → P(T ) be a carrier such that C(σ) is contractible for all σ ∈ K.Then:

1. If f, g : ‖K‖ → T are both carried by C, then f and g are homotopic.

2. There exists a mapping f : ‖K‖ → T carried by C.

Proof. Let C be a carrier from K, such that C(σ) is contractible for allσ ∈ K.

We will start by proving (1). Suppose that f, g : ‖K‖ → T are bothcarried by C, the we will construct a homotopy F : ‖K‖ × I → T . Nowwe have that f(‖{v}‖) ∈ C({v}) and g(‖{v}‖) ∈ C({v}) for all v ∈ K0.Now we can consider ‖K‖× I to be a collection of cells of the form ‖σ‖× I,for σ ∈ K, and a function on ‖K‖ × I is continuous if and only if it isconinuous on each ‖σ‖ × I. Now it is clear that the boundry of ‖{v}‖ × Iis given by ‖{v}‖ × {0} ∪ ‖{v}‖ × {1}. Further let F (‖{v}‖, 0) = f(‖{v}‖),and F (‖{v}‖, 1) = g(‖{v}‖), and now by theorem 4.13 we can continuouslyextend F : ‖{v}‖× {0, 1} → T to ‖{v}‖× I. Now suppose that F is definedand continuous on ‖K‖ × {0, 1} ∪ ‖K≤n‖ × I, with K≤n =

⋃ni=−1Ki, and

that F (‖σ‖×I) ⊂ C(σ). Then take τ ∈ Kn+1, then the boundary of ‖τ‖×Iis in ‖K‖ × {0, 1} ∪ ‖K≤n‖ × I, and since for all proper faces σ of τ thenC(σ) ⊂ C(τ), it is clear that F taken on the boundry of τ × I is continuousand a subset of C(τ) which is contractible and hence, F can be extendedcontinuously to ‖τ‖ × I, and per induction we are now done.

We will now prove (2) by inductively constructing f as a continuousfunction. For {v} ∈ K0 then f(‖{v}‖) should be any point in C({v}).Suppose now that f is continuously defined on ‖K≤n‖, with f(‖σ‖) ⊂ C(σ),

29


for all σ ∈ K≤n. Now take τ ∈ Kn+1, then for each simlex σ of τ we havethat

f(‖σ‖) ⊂ C(σ) ⊂ C(τ) (4.3)

and hence f on the boundary of τ is a subset of C(τ), so since the realizationof a simplex is a cell we can now extend f continuously to ‖τ‖, in such away so f(‖τ‖) ⊂ C(τ). And since f is continuous on K, if and only if it iscontinuous on each simplex of K, we are done.

The next theorem will be actually be needed for proving the duality,which is the goal of this chapter, but also to prove some further theorems.

Theorem 4.15. Let K be a simplicial complex, let P be a finite poset and letf : K → ∆(P ) a simplicial map. Define P≥x to be all chains in P consistingof elements all greater than equal to x. Suppose that ‖f−1(P≥x)‖ ⊂ ‖K‖,for all x ∈ P , is contractible. Then f induces homotopy equivalence between‖K‖ and ‖∆(P )‖.

Proof. Suppose that f−1(P≥x) are contractible, then define

C : ∆(P )→ P(‖K‖) (4.4)

given by C(σ) = ‖f−1(P≥minσ)‖. By part two of lemma 4.14 then thereis a continuous g carried by C. Now let C ′ : ∆(P ) → P(‖∆(P )‖) givenby C ′(σ) = ‖P≥minσ‖. Now it is clear that C ′(σ) is contractible for allσ ∈ ∆(P ) since it is all simplices of ∆(P ) with elements greater than orequal to minσ. It is clear that f ◦ g is carried by C ′ since g is carried byC, and f ◦ C is C ′, and trivially likewise the identity map on ‖∆(P )‖ iscarried, and hence by lemma 4.14 2 we get that f ◦ g is homotopic to theidentity. Let C ′′ : ∆(P ) → P(‖K‖) given as follows: For σ ∈ K thenC ′′(σ) = ‖f−1(P≥min f(σ))‖, then C ′′(σ) is contractible by assumption andclearly it carries g ◦ f since f(σ) ⊂ ‖P≥min f(σ)‖ and g is carried by C.Likewise the identity map on ‖K‖ is carried by C ′′ and hence by lemma4.14 g ◦ f is homotopic to the identity and therefore ‖∆(P )‖ is homotopicequivalent to ‖K‖.

Theorem 4.16. Let K be a simplicial complex and let (Ki)i∈I , be a finitefamily of simplicial complexes such that K =

⋃i∈I Ki. Suppose that every

nonempty finite intersection Ki1 ∩ . . . ∩ Kin is contractible, then ‖K‖ and‖N (Ki)‖ are homotopic equivalent.

Proof. Let f : K → N (Ki), where K,N (Ki) are both viewed as posetsordered by subset-ordering; and for σ ∈ K let f(σ) = {i ∈ I|σ ∈ Ki}. Thenf ′ : ∆(K)→ N (Ki) given as follows: For {σ0, . . . , σn} ∈ ∆(K) then

f ′({σ0, . . . , σn}) = {f(σ0), . . . , f(σn)} (4.5)

30

4.3. DUALITY

is well defined and simplicial. Now take arbitrary τ ∈ N (Ki) viewed as asubset. Then f ′−1(P≥τ ) =

⋂i∈τ Ki since elements that are mapped to τ or

greater must at least be in all the same Ki as τ . But now the result followsfrom theorem 4.15 and the assumption.

Corollary 4.17. If C is a crosscut of a poset P , then ‖Γ(P,C)‖ and ‖∆(P )‖are homotopic equivalent.

Proof. For any x ∈ C let Kx := ∆(P≤x∪P≥x), then it is clear that (Kx)x∈Cis a covering of ∆(P ), i.e., ∆(P ) =

⋃x∈C Kx since every element in P is

comparable to some element in C. If we take A ⊂ C such that⋂x∈AKx 6= ∅,

that means there is either an element p1 ∈ P such that p1 ≥ k for all k ∈ A,or there is p0 ∈ P such that p0 ≤ k for all k ∈ A. If p1 exists there isa least such we will call this element p1, since C is a crosscut, and hence⋂x∈AKx is contractible because it consists of the vertices with p1 as the

least element in each. If p0 exists there is a p0 which is the greatest elementsmaller than all k ∈ A, and then

⋂x∈AKx is contractible since it consists

of all simpleces with p0 as the greatest element. Now we need to show thatΓ(P,C) = N (Kx), but this is clear since A ⊂ C has

⋂x∈AKx 6= ∅ if and

only if it either has a lower or greater bound, and hence we are done.

4.3 Duality

This section is inspired by [2] where the 4.18 is stated. I have followed thesame proof-strategy as in [2], but the proof is deviced by me.

4.3.1 Preliminaries

In this section we will finally be able to state and prove the second dualityof this thesis. So let K be a simplicial complex on the vertex set V , then letC be the set of subsetwise least nonfaces of K, meaning that

C := min(D[V ]−K) (4.6)

, and let KΓ be the simplicial complex such that {c0, . . . , cn} ∈ KΓ forc0, . . . , cn ∈ C, whenever c0 ∪ . . . ∪ cn 6= V . This is clearly a simplicialcomplex since if c0 ∪ . . . ∪ cn 6= V , then no subset of {c0, . . . , cn} will haveunion equal to V . Now it is clear that this complex is not the same as theAlexander dual complex.

Now we can state the that theorem we want to prove:

Theorem 4.18. Let K be a simplicial complex on the vertex set V , with|V | − 1 = N , then Hi(K) ∼= Hj(KΓ) whenever i+ j = N − 2.

It is clear that this is somehow related to the Alexander duality sincethe two statements are almost identical. The only difference is that here

31

4.3. DUALITY

we have replaced K∨ with KΓ. And that is actually the heart of the proof:Since the reduced cohomology is stable under homotopy equivalences, thenwe wish to show that there is a homotopy equivalence between ‖K∨‖ and‖KΓ‖, and for this we will need the theorems from the previous subsection.

4.3.2 Proof

Proof of theorem 4.18 . Let ∂D[V ] := D[V ] − V and sdK∨ := ∆(K∨ − ∅),where K∨ − ∅ is viewed as a poset ordered by the subset-ordering. Nowdefine

f : sdK∨0 → ∂D[V ]−K (4.7)

such that for any σ ∈ ∆(K∨ − ∅)0 then f(σ) = V − σ. This is well definedsince σ 6= ∅ and since σ ∈ K∨−∅ and hence V − σ /∈ K. Now we see that if{σ0, . . . , σn} ∈ sdK∨, then σ0 ⊂ . . . ⊂ σn, but then f(σn) ⊂ . . . ⊂ f(σ0), so{f(σn), . . . , f(σ0)} ∈ ∆(∂D[V ]−K). Now take an arbitrary x ∈ ∂D[V ]−Kthen f−1((D[V ] − K)≥x) consists of all simplices in sdK∨ where V − x isgreater or equal to all elements in the simplex, and hence this is contractible.We can now by theorem 4.15 conclude that ‖sdK∨‖ and ‖∆(∂D[V ] −K)‖are homotopic equivalent.

Now let C be given as the set of minimal non-faces of K. We now whishto show that C is a crosscut of ∂D[V ]−K. It is clear that since the elementsof C are minimal no two distinct elements are related. It is clear that givenan arbitrary element p in ∂D[V ]−K there must always be some element inC such that it is smaller than p, since C consists of minimal elements. Givenan arbitrary nonempty subset A of C then there can’t be a lower bound forA in ∂D[V ] − K since C concists of minimal elements. It is clear that if⋃c∈A c 6= V , then A has an upper bound, namely

⋃c∈A c ∈ ∂D[V ] −K, if⋃

c∈A c = V , then A is unbounded. We now have that C is a crosscut of∂D[V ]−K. And it is clear that Γ(∂D[V ]−K,C) = KΓ, and hence we nowhave that ‖∆(∂D[V ]−K)‖ is homotopic to ‖KΓ‖.

And hence we get by use of the Alexander duality that

Hi(K) ∼= Hj(K∨) ∼= Hj(sdK∨) ∼= Hj(∆(∂D[V ]−K)) ∼= Hj(KΓ) (4.8)

whenever i+ j = N − 2.

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Bibliography

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[2] Bjorner, Anders; Butler, Lynne M.; Matveev, Andrej O. ”Note on aCombinatorial Application of Alexander Duality”. In: Journal of Com-binatorial Theory, Series A, vol 80 (1997), pp. 163-165.

[3] Bjorner, Anders; Tancer, Martin: ”Note: Combinatorial Alexander Du-ality - A Short and Elementary Proof”. In: Discrete & ComputationalGeometry, vol 42 (2009), pp. 586-593.

[4] Hatcher, Allen: Algebraic Topology. Cambridge University Press: NewYork 2010.

[5] Johnson, Jacob: Simplicial Complexes of Graphs. Springer-Verlag:Berlin 2008.

[6] Munkres, James R.: Elements of Algebraic Topology. The Ben-jamin/Cummings Publishing Company, Inc: Menlo Park 1984.

[7] Munkres, James R.: Topology - Internation edition - Second edition.Prentice Hall: Upper Addle River 2000.

[8] Stanley, Richard P.: Enumerative Combinatorics - Second edition ver-sion of 9 May 2011. http://www-math.mit.edu/∼rstan/ec/ec1 visited10th of may 2011.

[9] Walker, James W. ”Homotopy Type and Euler Characteristic of Par-tially Ordered Sets”. In: European Journal of Combinatorics, vol 2(1981), pp. 373-384.

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duality theorems for simplicial complexes

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